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Moore-Penrose Pseudo-Inverse Matrix

by Angharad (Pilgrim)
on Aug 01, 2005 at 18:55 UTC ( #480010=perlquestion: print w/replies, xml ) Need Help??

Angharad has asked for the wisdom of the Perl Monks concerning the following question:

I'm trying to write some code to calculate the Moore-Penrose Pseudo-inverse Matrix of a matrix (strangely enough :) ).
B is the pseudo-inverse of A if the following terms are true.
A.B.A = A B.A.B = B A(transposed) = A B(transposed) = B
I was wondering what the best way to tackle this might be. I'm familisising myself with PDL which is good because it allows me to transpose matrices very easily. I was thinking in terms of something like this
($flag) = Inverseinv(A) # A being a matrix if(Inversepinv == 1) { B = A # where B is the transpose of A }
The Inverseinv function would then do the tests to see if B does equal A as set up above.
As someone relatively inexperienced, I am just asking if this is the best way to go about solving this task and if not, any suggestions as to how else I could tackle it would be much appreciated. I'm not necessarily asking for code by the way, just suggestions, but if you have a snippet or two, that would obviously be appreciated too.
Thanks in advance.

Replies are listed 'Best First'.
Re: Moore-Penrose Pseudo-Inverse Matrix
by chas (Priest) on Aug 01, 2005 at 20:11 UTC
    My recollection (which may be in error) is that the Moore-Penrose inverse of A is: (A^tA)^{-1}A^t which exists if A^tA is invertible (i.e. if the columns of A are independent.) (I'm using ^ to indicate exponenents as in TeX source.) A^tA is square, and there are many existing routines to invert a square matrix - I'm sure there are some Perl modules that do this, although you could write your own as an exercise. You indicate above that A and the M-P inverse are symmetric, but I don't believe this is correct. The point of the M-P inverse is that it can be applied to non-square matrices.
    chas
    (Update: BTW, if you have a system of equations Ax=b then x=(A^tA)^{-1}A^tb is the usual "least squares" solution, i.e. the x for which Ax is closest to b.)
      Thanks for the suggestions :)

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